# Equivalence of Definitions of Inverse Mapping

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## Theorem

Let $S$ and $T$ be sets.

The following definitions of the concept of **Inverse Mapping** are equivalent:

### Definition 1

Let $f: S \to T$ be a mapping.

Let $f^{-1} \subseteq T \times S$ be the inverse of $f$:

- $f^{-1} := \set {\tuple {t, s}: \map f s = t}$

Let $f^{-1}$ itself be a mapping:

- $\forall y \in T: \tuple {y, x_1} \in f^{-1} \land \tuple {y, x_2} \in f^{-1} \implies x_1 = x_2$

and

- $\forall y \in T: \exists x \in S: \tuple {y, x} \in f$

Then $f^{-1}$ is called the **inverse mapping of $f$**.

### Definition 2

Let $f: S \to T$ and $g: T \to S$ be mappings.

Let:

- $g \circ f = I_S$
- $f \circ g = I_T$

where:

- $g \circ f$ and $f \circ g$ denotes the composition of $f$ with $g$ in either order
- $I_S$ and $I_T$ denote the identity mappings on $S$ and $T$ respectively.

That is, $f$ and $g$ are both left inverse mappings and right inverse mappings of each other.

Then:

- $g$ is
**the inverse (mapping) of $f$** - $f$ is
**the inverse (mapping) of $g$**.

## Proof

### Definition 1 implies Definition 2

Let $f^{-1}: T \to S$ be an inverse mapping of $f: S \to T$ by definition 1.

From Mapping is Injection and Surjection iff Inverse is Mapping it follows that $f^{-1}$ is a bijection.

By Composite of Bijection with Inverse is Identity Mapping:

- $f^{-1} \circ f = I_S$
- $f \circ f^{-1} = I_T$

Thus $f^{-1}: T \to S$ is an inverse mapping of $f: S \to T$ by definition 2.

$\Box$

### Definition 2 implies Definition 1

Let $g: T \to S$ be an inverse mapping of $f: S \to T$ by definition 2.

From Left and Right Inverse Mappings Implies Bijection it follows that both $f$ and $g$ are bijections.

From Mapping is Injection and Surjection iff Inverse is Mapping it follows that $g = f^{-1}$.

Thus $g: T \to S$ is an inverse mapping of $f: S \to T$ by definition 1.

$\blacksquare$

## Sources

- 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): $\S 3.6$. Products of bijective mappings. Permutations